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TECHNIQUES OF DC CIRCUIT ANALYSIS:
Application to operational amplifier circuit
SKEE 1023
1
Operational amplifier: Integrated circuit consisting several
transistors and resistors normally used for analog circuit design.
Offset null(5)
Inverting input (2)
Non-inverting input (3)
−
Positive supply Vcc+ (7)
Output (6)
+
Offset null(1)
Negative supply Vcc− (4)
2
Operational amplifier: Integrated circuit consisting several
transistors and resistors normally used for analog circuit design.
How can we analyse circuit containing op-amps (consistings of
hundreds of transistors and resistors!) using the knowledge that we
have so far?
The answer is modeling !
An op-amp can be modeled by a very good voltage
amplifier circuit
Later we’ll see that the analysis is even simplified
when we assumed an ideal op-amp.
3
vo = Avd = A(v2 - v1)
vo - output voltage with respect
to ground
v1
A -
−
vd
+
v2
Ro
Ri
+
−
Avd
v0
open loop gain
v1 - inverting input voltage with
respect to ground
v2 - non-inverting input voltage with
respect to ground
An op-amp can be modeled by a very good voltage
amplifier circuit
4
An op-amp can be modeled by a very good voltage
amplifier circuit
vo = Avd = A(v2 - v1)
vo - output voltage with respect
to ground
v1
A -
−
vd
+
v2
Ro
Ri
+
−
open loop gain
v1 - inverting input voltage with
respect to ground
v0
v2 - non-inverting input voltage with
respect to ground
Avd
vo
Positive saturation
Vcc
-Vcc/A
Vcc/A
vd
Negative saturation
-Vcc
5
Let’s look at a practical circuit: Unity gain buffer
+
−
Vcc
+
Vs
+
−
−
+
• In drawing op-amp circuits, we
normally do not include power
supplies
vo
−
+
Vcc
−
6
Let’s look at a practical circuit: Unity gain buffer
+
Vs
+
−
−
+
• In drawing op-amp circuits, we
normally do not include power
supplies
vo
−
7
Let’s look at a practical circuit: Unity gain buffer
Using KVL, we can write:
Vs = iRi + iRo + Aovin
Vo = iRo + Aovin
i
Vin = iRi
+
Vs
+
−
+
Ro
R
vin i
−
+
− Aovin
−

Vo

Vs 1 
+
vo
1
Ri
Ro  Ao Ri
For Ri >> Ro
Vo
1

Vs 1  1
Ao
−
For Ao>>1
Whenever vin 0 , Aovin increases until
vin=0 and hence the current cease to flow.
Vo
1
Vs
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Modeling of an ideal op-amp
1. In previous analysis we have assumed Ri>>Ro
In practical op-amps it is true that Ri >> Ro and for ideal op-amp we will
assume Ri → and Ro → 0
2. In previous analysis we have assumed Ao>> 1
In practical op-amps it is true that Ao>> 1 and for ideal op-amp we will
assume Ao →
As a result of 1 and 2, when analysing an op-amp circuit with feedback, we will
assume that:
i. i1 = 0, and i2 = 0
ii. v2 –v1 = 0
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Let’s look back at the buffer circuit using ideal op-amp
Since vo is tied to v1 and in ideal op-amp
v1=v2, it is obvious that vo = Vs (as we
have seen before)
v2 +
Vs
+
−
v1
−
+
vo
−
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Eg.2 (PP 5.2)
Find the closed-loop gain vo/vs. Determine current i when vs = 2V
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OP AMP circuits
In SKEE1023, we will study some of the important op amp circuits:
•
•
•
•
Inverting amplifier
Non-inverting amplifier
Summing amplifier
Difference amplifier
You should :
• Know how (not memorize!) to derive the input-output
relationship (Gain = vo/vi))
• Able to identify the circuit correctly
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Inverting Amplifier
Reverses the polarity of a signal while amplifying it
Since i1 = i2, and v1 = v2 = 0
Rf
vi  0 0  vo

 vo  
vi
R1
Rf
Ri
13
Non-inverting Amplifier
Maintains the polarity of a signal while amplifying it
Since i1 = i2, and v1 = v2 = vi

vi vi  vo

R1
Rf
 1 
 1
1 
  vi
 vo 
 
 Rf 
 Rf R1 
 Rf 
vi
 vo  1 
R1 

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Summing Amplifier
Output is the inverted, scaled sum of the voltages applied to the input
KCL at node a gives: i = i1 + i2 + i3
And, since v+ = v- = 0
v o v1 v 2 v 3



Rf R1 R2 R3



R

R
R
v o   f v 1  f v 2  f v 3 
R2
R3 
 R1
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Difference Amplifier
Output is proportional to the difference between the two inputs voltages
Using voltage division rule,
 R4 
v 2
v b  
 R4  R3 
With va = vb, applying KCL at the inverting input,
v1  v b v b  v o

R1
R2
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Difference Amplifier
Output is proportional to the difference between the two inputs voltages
It can be shown that:
If
R1 R3

R2 R 4
R2 1  R1 

R2 
 R2 
vo 
v 2   v1
R
 R1 
R1 1  3 

R4 

vo 
R2
v 2  v1 
R1
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